The displacement field D is related to the vacuum permittivity and polarization by which relation?

The displacement field D is related to the vacuum permittivity and polarization by which relation?

• A. D = ε0 E
• B. D = ε0 E + P
• C. D = ε E
• D. D = P

Answer: (B)

The displacement field D is the field that combines the driving electric field with the material’s response through polarization. It is defined as D = ε0 E + P, where ε0 E represents the vacuum response and P is the polarization (the dipole moment density within the material). This definition is why Gauss’s law in matter is written as ∇·D = ρ_free, separating free charges from bound charges contained in P. In vacuum, there is no polarization, so P = 0 and D reduces to D = ε0 E. In a linear dielectric, polarization relates to the field as P = ε0 χe E, giving D = ε0 E + ε0 χe E = ε0 (1 + χe) E = ε0 εr E. The other forms aren’t the general definition. D = ε0 E is only the vacuum case, while D = ε E would require introducing a different ε that hides the P term, and D = P neglects the field contribution entirely.

The displacement field D is the field that combines the driving electric field with the material’s response through polarization. It is defined as D = ε0 E + P, where ε0 E represents the vacuum response and P is the polarization (the dipole moment density within the material). This definition is why Gauss’s law in matter is written as ∇·D = ρ_free, separating free charges from bound charges contained in P.

In vacuum, there is no polarization, so P = 0 and D reduces to D = ε0 E. In a linear dielectric, polarization relates to the field as P = ε0 χe E, giving D = ε0 E + ε0 χe E = ε0 (1 + χe) E = ε0 εr E.

The other forms aren’t the general definition. D = ε0 E is only the vacuum case, while D = ε E would require introducing a different ε that hides the P term, and D = P neglects the field contribution entirely.